A proton with an initial speed of 970,000 m/s is brought to rest by an electric field.?

A proton with an initial speed of 970,000 m/s is brought to rest by an electric field. (a) Did the proton move into a region of higher potential or lower potential? higher lower (b) What was the potential difference that stopped the proton? V (c) What was the initial kinetic energy of the proton

Answer:

Protons move by themselves from higher to lower potential.

Since the proton is stopped by the potential, it was moving toward a high potential.
b)

1/2 mv^2 = V*e
m =1.67 × 10-27 kg
e =1.602 e-19 C

V = 1/2 mv^2 /e = 0.5*1.67e-27*970000*970000/1.602 e-19

V = 4904 V

c)
1/2 mv^2 = V*e
Hence
its initial k.e was 4904 [eV]

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